Let `P(x, y)` be a point on AB such that
`" "PA=2PB`
`rArr" "(PA)/(PB)=(2)/(1)" "rArr" "PA:PB=2:1`
So, by using section formula for internal division,
`" "x=(mx_(2)+nx_(1))/(m+n), y= (my_(2)+ny_(1))/(m+n)`
`rArr" "x=(2(-3)+1(5))/(2+1)rArr" "x=-(1)/(3)`
and `" "y=(2(2)+1(-4))/(2+1)rArr" "y=0`
`rArr" "` Required point `-=(-(1)/(3), 0)`
`" ""BUT"" "`
This is not the end of this question.
Think : Is it not possible that `P(x,y)` divides AB externally in the ratio 2 : 1.
So, by using section formula for external division,
`therefore" "x=(2(-3)-1(5))/(2-1)rArr" "x=-(11)/(1)rArr" "x=-11`
and `" "y=(2(2)-1(-4))/(2-1)rArr" "y=(8)/(1)rArr" "y=8`
So, co-ordinates of `p` are (-11, 8) also.
Hence, required points are `(-(1)/(3),0) and (-11, 8)`.
